[hihoCoder] 区域周长 解题报告 - Jonah的专栏 - 博客频道 - CSDN.NET


[hihoCoder] 区域周长 解题报告 - Jonah的专栏 - 博客频道 - CSDN.NET
给定一个包含 N × M 个单位正方形的矩阵,矩阵中每个正方形上都写有一个数字。
对于两个单位正方形 a 和 b ,如果 a 和 b 有一条共同的边,并且它们的数字相等,那么 a 和 b 是相连的。
相连还具有传递性,如果 a 和 b 相连,b 和 c 相连,那么 a 和 c 也相连。
给定一个单位正方形 ss 和与 s 相连的所有单位正方形会组成一个区域 R 。小Hi想知道 R 的周长是多少?

输入

第一行包含4个整数 N , M ,x 和 y , N 和 M 是矩阵的大小, x 和 y 是给定的单位正方形 s 的坐标。(1 ≤ N , M ≤ 100, 0 ≤ x < N , 0 ≤ y < M )
以下是一个 N × M 的矩阵 AAij 表示相应的正方形上的数字。(0 ≤ Aij ≤ 100)

输出

输出一个整数表示 R 的周长。

样例输入

6 5 2 1
0 0 1 2 2
3 1 1 3 7
4 3 1 3 7
4 3 0 3 2
4 3 3 0 1
4 5 5 5 5
样例输出
10

思路:求相邻正方形连成的周长,基本思路还是DFS.每找到一个未被访问过相邻的正方形周长+2,如果找到一个被访问过的正方形,周长-2.
在写代码的时候为了防止从A->B, 再从B->A, 我设置了一个数组来记录已经出现过的相邻的正方形.但其实没必要,只要在做DFS的时候判断一下不要回到上一个调用当前正方形的地方就行了.

int N, M, x, y,tem, ans=2;
vector<vector<int> > map;
vector<vector<set<int> > > neigh;

void DFS(int x, int y,int preX, int preY, int val)
{
    if(x<0 || x>=N || y<0 || y>=M ||map[x][y]!=val||neigh[x][y].count(preX*M+preY)) return;
    if(neigh[x][y].size() == 0) ans += 2;
    else ans -= 2;
    neigh[preX][preY].insert(x*M+y);
    neigh[x][y].insert(preX*M+preY);
    DFS( x+1, y, x, y, val);
    DFS( x-1, y, x, y, val);
    DFS( x, y+1, x, y, val);
    DFS( x, y-1, x, y, val);
}

int main()
{
    cin >> N >> M >> x >> y;
    map.resize(N, vector<int>(M));
    neigh.resize(N, vector<set<int> >(M));
    for(int i = 0; i < N; i++)
        for(int j =0; j < M; j++)
        {
            cin >> tem;
            map[i][j] = tem;
        }
    DFS( x, y, x, y, map[x][y]);
    cout << ans << endl;
    return 0;
}
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